Force of friction. Non conservation of energy

[anna sung]

Homework Statement

A 15 kg child slides, from rest, down a playground slide tht is 4.0 m long, as shown in the figure. The slide makes a 40 degrees angle with the horizontal. The child's speed at the bottom is 3.2 m/s. What was the force of friction that the sldie was exerting on the child?

Ok. The answer in the texbook says 75N.

Homework Equations

Wnc=Efinal-Einitial

The Attempt at a Solution

Here is my WROng attempt

work=Kinetic Energy 2 - Kinetic Energy 1
work = 1/2 m (V2squared - v1squared)
w=1/2 (15 kg) (3.2 m/s squared - 0)
w=7.5 (10.24 m/s)
w=76.8 J

then I..

Work=force (distance) (cos40)
f= w/d (cos 40)
f=76.8 J / 4.0 (cos 40)
f=14.7 N
could you please actually plug in the numbers and show me how to get to the answer?
rather than giving just the steps? please thank you!

 

[PhanthomJay]

We can assist you in your attempt to get the answer, but we don't provide the answer.
The goal is for you to provide the answer, based on assistance by others.

Homework Statement

A 15 kg child slides, from rest, down a playground slide tht is 4.0 m long, as shown in the figure. The slide makes a 40 degrees angle with the horizontal. The child's speed at the bottom is 3.2 m/s. What was the force of friction that the sldie was exerting on the child?

Ok. The answer in the texbook says 75N.

Homework Equations

Wnc=Efinal-Einitial

You have to define what you mean by Einitial and Efinal. Einitial and Efinal each include both kinetic and potential energies.

The Attempt at a Solution

Here is my WROng attempt

work=Kinetic Energy 2 - Kinetic Energy 1

Again , be careful with this equation in the definition of 'Work". This equation is for total (or also called 'net') work, where total work includes the work done by both conservative and non-conservative forces.

then I..

Work=force (distance) (cos40)

When you are trying to find the force of friction based on the work done by friction, you should note that since the friction force and displacement are along the same axis, then theta is not 40 degree in the W =(f)(d)(cos theta) equation.

could you please actually plug in the numbers and show me how to get to the answer?
rather than giving just the steps? please thank you!

Please try again using the 'Wnc = Ef -Ei' equation you tried above, but make the correction for the Ef and Ei terms.
Once you get the correct Wnc value, where Wnc is the work done by friction in this case, then use the correct W =fdcostheta equation to solve for the friction force.